Show the divergence of $\sum_{k=2}^\infty \frac{1}{\sqrt k (\ln k)^2} $

Question

I'm attempting to understand how the following sum behaves for different values of $p$ and $q$: $$\sum_{k=2}^\infty \frac{1}{ k^q (\ln k)^p} $$I think I have figured out every case, except when $q \in (0,1)$ and $p>1. $ To determine what to do, I gave concrete values to $p$ and $q$. Namely, $q=\frac{1}{2}$, and $p=2$. But I still can't figure out how to prove the behavior of the series. Using Wolfram Alpha, I know that the series below diverges, but I don't know how to prove it. Does anyone have any insight to showing either the general case above or the more concrete case below? Thanks in advance.

$$\sum_{k=2}^\infty \frac{1}{\sqrt k (\ln k)^2} $$

Answer 1

Informally:

Actually, $\ln k$ grows slower than any positive power of $k$: for any $\alpha>0$ there is an $n$ such that

$$k>n\implies\ln k<k^\alpha$$

as, taking the derivative, $k^{-1}$ grows slower than $k^{\alpha-1}$.

Then you can replace $q$ by $q+\alpha<1$ and drop the $\ln$ factor, and the resulting series is bounded below by the harmonic one.

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More about this:

In a way, $\ln k=k^0$. Indeed,

$$\ln k=\lim_{\alpha\to0}\frac{k^\alpha-1}\alpha.$$

Answer 2

For the general case, use the integral test for series. Observe that $$\sum \frac{1}{k^p\ln(k)^q} \approx \int_1^{\infty}\frac{dx}{x^p\ln(x)^q} = \int_0^{\infty}\frac{e^{-(p-1)u}}{u^q}du$$ It should be easier to deduce what values of $p$ and $q$ lead to convergence.

Answer 3

You can prove the following: $\left(\ln k\right)^2 < \sqrt[3]{k}$ for large enough $k \ge 2$. And this amounts to: $\ln k < k^{\frac{1}{6}}$. The ratio $\dfrac{k^{1/6}}{\ln k} \to +\infty$ as $k \to \infty$. Thus starting at $k_0$ on ward, $k^{1/6} > \ln k \implies \ln k - k^{1/6} < 0, k \ge k_0$.

Answer 4

From Cauchy's Condensation Test, the series $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}\log^2(n)}$ is convergent if anf only if the series $\sum_{n=1}^\infty \frac{2^{n/2}}{n^2\log^2(2)}$ converges. Since the latter series diverges (the general terms do not approach $0$), the series of interest diverges likewise.

Answer 5

The key fact is that $\dfrac{\ln x}{x^c} \to 0 $ as $x \to \infty$ for any $c > 0$.

So show this, start with $\dfrac{\ln x}{x} \to 0 $ as $x \to \infty$. Then $\dfrac{\ln x^c}{x^c} \to 0 $ as $x \to \infty$. But $\dfrac{\ln x^c}{x^c} =\dfrac{c\ln x}{x^c} \to 0 $.

Here is an easy way to show that $\dfrac{\ln x}{x} \to 0 $.

$\ln(x) =\int_1^x \dfrac{dt}{t} \le\int_1^x \dfrac{dt}{t^{1/2}} =\dfrac{x^{1/2}-1}{1/2} \lt 2x^{1/2} $ so $\dfrac{\ln(x)}{x} \le \dfrac{2}{x^{1/2}} \to 0 $.

You can also show it directly by a modification of this last proof.

For any $c > d > 0$ and $x > 1$,

$\begin{array}\\ \ln(x) &=\int_1^x \dfrac{dt}{t}\\ &\lt\int_1^x \dfrac{dt}{t^{1-d}} \qquad\text{since } t > t^{1-d}\\ &=\dfrac{x^{d}-1}{d}\\ &<\dfrac{x^{d}-1}{d}\\ \text{so}\\ \dfrac{\ln(x)}{x^c} &\le \dfrac{x^{d-c}}{d}\\ &= \dfrac{1}{dx^{c-d}}\\ &\to 0\\ \end{array} $

Answer 6

At the original time that I posted this, many of these answers did not make complete sense to me, so I figured that I would post my own answer to this question that would have made sense to me back then.

We will solve the general case, when $q \in (0,1)$ and $p>1$. To solve this problem, we will have to use both the direct comparison test and the integral test.

Let $\displaystyle a_k = \frac{1}{k^q (\ln k)^p}$ and $\displaystyle b_k = \frac{1}{k\ln k}$. Arguably, the hardest part of this problem is understanding that for sufficiently large $k$, $a_k > b_k$. To prove this, we take the limit of their quotient.

$$\lim_{k \to \infty} \frac{k \ln k}{k^q (\ln k)^p} = \lim_{k \to \infty} \frac{k^{1-q}}{(\ln k)^{p-1}}$$

Because $k^\alpha \gg (\ln k)^\beta$ for any $\alpha,\beta>0$, we know that the above limit is equal to $\infty$. Therefore, there must exist an $N$ such that for $k>N$, $k\ln k > k^q (\ln k)^p$. Thus, we conclude that for sufficiently large $k$, $$k \ln k > k^q (\ln k)^p \iff \frac{1}{k\ln k} < \frac{1}{k^q (\ln k)^p} \iff b_k < a_k$$

Now that we have that out of the way, if we can prove that $\sum_{k=2}^\infty b_k$ diverges, then $\sum_{k=2}^\infty a_k$ must diverge as well by the direct comparison test.

To prove that $\sum_{k=2}^\infty b_k$ diverges, we use the integral test.

$$\begin{align*} \int_2^\infty \frac{1}{x \ln x} dx &= \lim_{b \to \infty} \int_2^b \frac{1}{x \ln x}dx, \qquad u = \ln x, \quad du = \frac{1}{x}dx \\ &=\lim_{b \to \infty} \int_{\ln 2}^{\ln b} \frac{1}{u}du \\ &= \lim_{b \to \infty} \Big[\ln|u|\Big]_{\ln2}^{\ln b}\\ &= \lim_{b \to \infty} \Big(\ln\big|\ln b\big| - \ln\big|\ln2\big|\Big)\\ &=\infty \end{align*}$$

Since the improper integral diverges, the associated sum $\sum b_k$ must diverge as well from the integral test. Therefore, we may conclude that $\displaystyle \sum_{k=2}^\infty \frac{1}{k^q (\ln k)^p}$ diverges for $q\in (0,1)$ and $p >1$.